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3 years ago

Acceleration involves a change in speed or what?

Physics

1 answer:

jolli1 [7]3 years ago

5 0

Answer:

Velocity would be your answer.

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In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s

statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

$F = ma$

m = mass

a = Acceleration

By Hooke's law force is described as

$F = k\Delta x$

Here,

k = Gravitational constant

x = Displacement

To develop this problem it is necessary to consider the two cases that give us concerning the elongation of the body.

The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

$k\Delta x = mg$

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

$k (9.5-l)=0.2*g$

$k (9.5-l)=0.2*9.8$

For state 2 we have that with 1Kg there is an elongation of 12cm

$k (12-l)= 1*g$

$k (12-l)= 1*9.8$

We have two equations with two unknowns therefore solving for both,

$k = 3.136N/cm$

$l = 8.877cm$

In this way converting the units,

$k = 3.136N/cm(\frac{100cm}{1m})$

$k = 313.6N/m$

Therefore the spring constant is 313.6N/m

3 0

3 years ago

Question 18 of 20

disa [49]

Answer:D

Explanation:

5 0

3 years ago

Twins Jody and Taylor are rearranging the furniture in their bedroom and want to move a dresser across the room. The dresser has

xeze [42]

Answer:

yes, They will be able to move the dresser.

Explanation:

sliding force 90N

55N + 38N = 93N

therefore, yes the twins can move the dresser

7 0

3 years ago

Calculate the net force on particle qi.

aksik [14]

Answer:

-12.12

Explanation:

i just added them together and it worked.

4 0

2 years ago

At the local swimming pool, the diving board is elevated h = 8.5 m above the pool surface and overhangs the pool edge by L = 2 m

Tasya [4]

Answer:

Part a)

$t_w = \sqrt{\frac{2h}{g}}$

Part b)

$t_w = 1.32 s$

Part c)

$d_w = 6.88 m$

Explanation:

Part a)

As we know that the diver will have zero velocity in vertical direction

so here we can say that

$\Delta y = v_y t + \frac{1}{2}a_y t^2$

$h = \frac{1}{2}gt^2$

$t_w = \sqrt{\frac{2h}{g}}$

Part b)

as we know that

$h = 8.5 m$

$g = 9.81 m/s^2$

so we will have

$t_w = \sqrt{\frac{2(8.5)}{9.81}}$

$t_w = 1.32 s$

Part c)

Distance covered by diver from the edge of the pool is given as

$d_w = L + v_w t_w$

$d_w = 2 + (3.7)(1.32)$

$d_w = 6.88 m$

7 0

4 years ago